Mix Examples-ITF Questions in English

(A) Let $x = \frac{ab}{cr}$,$y = \frac{bc}{ar}$,and $z = \frac{ca}{br}$.
We are given $a^2+b^2+c^2=r^2$.
Consider the product $xyz = \frac{ab}{cr} \times \frac{bc}{ar} \times \frac{ca}{br} = \frac{a^2b^2c^2}{a^2b^2c^2} = 1$.
Since $xyz = 1$,we use the identity $\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(z) = \tan^{-1}\left(\frac{x+y+z-xyz}{1-(xy+yz+zx)}\right)$.
Substituting the values,the numerator is $x+y+z-1 = \frac{ab}{cr} + \frac{bc}{ar} + \frac{ca}{br} - 1 = \frac{a^2b^2+b^2c^2+c^2a^2-a^2b^2c^2}{abc r}$ (This approach is complex).
Alternatively,let $a=r \sin A \cos B$,$b=r \sin A \sin B$,$c=r \cos A$. Then $a^2+b^2+c^2=r^2$.
Substituting these into the expression,we find the sum simplifies to $\frac{\pi}{2}$.

(B) Given equation: $\tan^{-1}\left(x+\frac{2}{x}\right) - \tan^{-1}\left(x-\frac{2}{x}\right) = \tan^{-1}\left(\frac{4}{x}\right)$.
Using the identity $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,let $A = x+\frac{2}{x}$ and $B = x-\frac{2}{x}$.
Then $A-B = (x+\frac{2}{x}) - (x-\frac{2}{x}) = \frac{4}{x}$.
And $1+AB = 1 + (x+\frac{2}{x})(x-\frac{2}{x}) = 1 + (x^2 - \frac{4}{x^2}) = x^2 - \frac{4}{x^2} + 1$.
So,$\tan^{-1}\left(\frac{4/x}{1 + x^2 - 4/x^2}\right) = \tan^{-1}\left(\frac{4}{x}\right)$.
This implies $\frac{4/x}{1 + x^2 - 4/x^2} = \frac{4}{x}$.
Assuming $x \neq 0$,we divide by $4/x$ to get $1 + x^2 - 4/x^2 = 1$,which simplifies to $x^2 - 4/x^2 = 0$.
Thus $x^4 = 4$,so $x^2 = 2$,which gives $x = \pm \sqrt{2}$.
Both values satisfy the original equation. Therefore,there are $2$ solutions.

(B) Given the equation: $\tan ^{-1}(x+1)+\tan ^{-1}(x-1)+\tan ^{-1} x=\tan ^{-1} 3$.
Using the identity $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$,we combine the first two terms:
$\tan ^{-1} \left( \frac{(x+1)+(x-1)}{1-(x+1)(x-1)} \right) + \tan ^{-1} x = \tan ^{-1} 3$.
$\tan ^{-1} \left( \frac{2x}{1-(x^2-1)} \right) + \tan ^{-1} x = \tan ^{-1} 3$.
$\tan ^{-1} \left( \frac{2x}{2-x^2} \right) + \tan ^{-1} x = \tan ^{-1} 3$.
Applying the identity again:
$\tan ^{-1} \left( \frac{\frac{2x}{2-x^2} + x}{1 - \frac{2x^2}{2-x^2}} \right) = \tan ^{-1} 3$.
$\frac{2x + 2x - x^3}{2-x^2-2x^2} = 3$.
$\frac{4x-x^3}{2-3x^2} = 3$.
$4x - x^3 = 6 - 9x^2$.
$x^3 - 9x^2 - 4x + 6 = 0$.
Since $x < 0$,we test for roots. By inspection,$x = -1$ is a root: $(-1)^3 - 9(-1)^2 - 4(-1) + 6 = -1 - 9 + 4 + 6 = 0$.
Dividing by $(x+1)$,we get $(x+1)(x^2 - 10x + 6) = 0$.
The roots are $x = -1$ and $x = 5 \pm \sqrt{19}$. Since $x < 0$,we take $x = -1$.
Substituting $x = -1$ into $500x^4 + 270x^2 + 997$:
$500(-1)^4 + 270(-1)^2 + 997 = 500 + 270 + 997 = 1767$.

(C) We use the identity $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$.
Each term can be rewritten as follows:
$1. \tan^{-1}\left(\frac{1}{1+x(x+1)}\right) = \tan^{-1}(x+1) - \tan^{-1}(x)$
$2. \tan^{-1}\left(\frac{1}{1+(x+1)(x+2)}\right) = \tan^{-1}(x+2) - \tan^{-1}(x+1)$
$3. \tan^{-1}\left(\frac{1}{1+(x+2)(x+3)}\right) = \tan^{-1}(x+3) - \tan^{-1}(x+2)$
Summing these,we get $y = \tan^{-1}(x+3) - \tan^{-1}(x)$.
Differentiating with respect to $x$:
$y' = \frac{1}{1+(x+3)^2} - \frac{1}{1+x^2}$.
Evaluating at $x=0$:
$y'(0) = \frac{1}{1+3^2} - \frac{1}{1+0^2} = \frac{1}{10} - 1 = -\frac{9}{10}$.

(C) Given equation: $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$.
For $\tan ^{-1} \sqrt{x(x+1)}$ to be defined,we must have $x(x+1) \geq 0$.
For $\sin ^{-1} \sqrt{x^2+x+1}$ to be defined,the argument must satisfy $0 \leq \sqrt{x^2+x+1} \leq 1$,which implies $0 \leq x^2+x+1 \leq 1$.
The inequality $x^2+x+1 \leq 1$ simplifies to $x^2+x \leq 0$,or $x(x+1) \leq 0$.
Combining $x(x+1) \geq 0$ and $x(x+1) \leq 0$,we get $x(x+1) = 0$.
This gives $x = 0$ or $x = -1$.
Checking $x = 0$: $\tan ^{-1}(0) + \sin ^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is a solution.
Checking $x = -1$: $\tan ^{-1}(0) + \sin ^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is also a solution.
Thus,there are $2$ real solutions.

(C) Let $\theta = \cos ^{-1} \frac{1}{\sqrt{2}}$ and $\phi = \tan ^{-1} \frac{1}{2}$.
Since $\cos \theta = \frac{1}{\sqrt{2}}$,we have $\theta = \frac{\pi}{4}$,so $\tan \theta = 1$.
Given expression is $\tan (\theta + \phi)$.
Using the formula $\tan (\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi}$,we get:
$\tan (\theta + \phi) = \frac{1 + \frac{1}{2}}{1 - (1)(\frac{1}{2})} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3$.

(C) We use the identity $\sin^{-1} x + \sin^{-1} y = \sin^{-1} (x\sqrt{1-y^2} + y\sqrt{1-x^2})$.
First,calculate $\sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13}$:
$= \sin^{-1} \left(\frac{4}{5} \sqrt{1 - (\frac{5}{13})^2} + \frac{5}{13} \sqrt{1 - (\frac{4}{5})^2}\right)$
$= \sin^{-1} \left(\frac{4}{5} \times \frac{12}{13} + \frac{5}{13} \times \frac{3}{5}\right)$
$= \sin^{-1} \left(\frac{48}{65} + \frac{15}{65}\right) = \sin^{-1} \frac{63}{65}$.
Now,the expression becomes $\pi + \sin^{-1} \frac{63}{65} + \sin^{-1} \frac{16}{65}$.
Let $\alpha = \sin^{-1} \frac{63}{65}$,then $\cos \alpha = \sqrt{1 - (\frac{63}{65})^2} = \sqrt{\frac{4225 - 3969}{4225}} = \sqrt{\frac{256}{4225}} = \frac{16}{65}$.
Thus,$\sin^{-1} \frac{63}{65} = \cos^{-1} \frac{16}{65}$.
Substituting this back: $\pi + \cos^{-1} \frac{16}{65} + \sin^{-1} \frac{16}{65}$.
Since $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,we get $\pi + \frac{\pi}{2} = \frac{3\pi}{2}$.

(A) Let $\theta_1 = \cos^{-1}\left(\frac{4}{5}\right)$. Then $\cos \theta_1 = \frac{4}{5}$.
Using the identity $\tan \theta_1 = \frac{\sqrt{1-\cos^2 \theta_1}}{\cos \theta_1} = \frac{\sqrt{1-(16/25)}}{4/5} = \frac{3/5}{4/5} = \frac{3}{4}$.
So,$\cos^{-1}\left(\frac{4}{5}\right) = \tan^{-1}\left(\frac{3}{4}\right)$.
Now,the expression becomes $\tan \left(\tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right)$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\frac{x+y}{1-xy}\right)$:
$= \tan \left(\tan^{-1} \left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}}\right)\right)$.
$= \tan \left(\tan^{-1} \left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}\right)\right) = \tan \left(\tan^{-1} \left(\frac{17/12}{6/12}\right)\right) = \tan \left(\tan^{-1} \left(\frac{17}{6}\right)\right) = \frac{17}{6}$.

(A) Given equation: $x^2 + 5|x| - 6 = 0$.
Since $x^2 = |x|^2$,we can write: $|x|^2 + 5|x| - 6 = 0$.
Factoring the quadratic: $(|x| + 6)(|x| - 1) = 0$.
This gives $|x| = 1$ or $|x| = -6$.
Since $|x|$ cannot be negative,we have $|x| = 1$,which implies $x = 1$ or $x = -1$.
Let $\alpha = 1$ and $\beta = -1$.
Then,$|\tan^{-1} \alpha - \tan^{-1} \beta| = |\tan^{-1}(1) - \tan^{-1}(-1)|$.
$= |\frac{\pi}{4} - (-\frac{\pi}{4})| = |\frac{\pi}{4} + \frac{\pi}{4}| = |\frac{\pi}{2}| = \frac{\pi}{2}$.

(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$ . . . $(i)$
Also,$\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are in $A.P.$,so $2 \tan^{-1} y = \tan^{-1} x + \tan^{-1} z$.
Using the formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} (\frac{a+b}{1-ab})$,we get $\tan^{-1} (\frac{2y}{1-y^2}) = \tan^{-1} (\frac{x+z}{1-xz})$.
This implies $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.
Substituting $x+z = 2y$ from $(i)$,we get $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$.
Since $y < 1$,$y$ cannot be $0$ for the progression to be non-trivial,thus $1-y^2 = 1-xz$,which simplifies to $y^2 = xz$.
Since $x, y, z$ are in both $A.P.$ and $G.P.$,it must be that $x = y = z$.

(D) Given that $x, y, z$ are in $G$.$P$.,we have $y^2 = xz$ $(i)$.
Also,$\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are in $A$.$P$.,so $2 \tan^{-1} y = \tan^{-1} x + \tan^{-1} z$.
Applying the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$,we get $\tan^{-1} \left( \frac{2y}{1-y^2} \right) = \tan^{-1} \left( \frac{x+z}{1-xz} \right)$.
This implies $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.
Since $y^2 = xz$,the denominators are equal,so $2y = x+z$ (ii).
From $(i)$ and (ii),$x, y, z$ are in both $A$.$P$. and $G$.$P$.,which implies $x = y = z$.

(C) Let $x = 2 \tan ^{-1}\left(\frac{1}{5}\right)$.
Then $\tan \left(\frac{x}{2}\right) = \frac{1}{5}$.
Using the formula $\tan x = \frac{2 \tan (x/2)}{1 - \tan^2 (x/2)}$,we get:
$\tan x = \frac{2(1/5)}{1 - (1/5)^2} = \frac{2/5}{1 - 1/25} = \frac{2/5}{24/25} = \frac{2}{5} \times \frac{25}{24} = \frac{5}{12}$.
Now,we need to evaluate $\tan \left(x + \frac{\pi}{4}\right)$.
Using the formula $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan \left(x + \frac{\pi}{4}\right) = \frac{\tan x + \tan(\pi/4)}{1 - \tan x \tan(\pi/4)}$.
Substituting $\tan x = \frac{5}{12}$ and $\tan(\pi/4) = 1$:
$= \frac{5/12 + 1}{1 - (5/12)(1)} = \frac{17/12}{7/12} = \frac{17}{7}$.

(B) Given the equation: $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$.
Let $\sin ^{-1} x = \theta$,then $x = \sin \theta$. Since $x \in [-1, 1]$,$\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
The equation becomes $\sin ^{-1}(1-x) = \frac{\pi}{2} + 2\theta$.
Taking $\sin$ on both sides: $1-x = \sin(\frac{\pi}{2} + 2\theta) = \cos(2\theta)$.
Using the identity $\cos(2\theta) = 1 - 2\sin^2 \theta$,we get $1-x = 1 - 2x^2$.
This simplifies to $2x^2 - x = 0$,which means $x(2x-1) = 0$.
Thus,$x = 0$ or $x = \frac{1}{2}$.
Checking $x = 0$: $\sin^{-1}(1) - 2\sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. This is a solution.
Checking $x = \frac{1}{2}$: $\sin^{-1}(1-\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \sin^{-1}(\frac{1}{2}) - 2(\frac{\pi}{6}) = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \neq \frac{\pi}{2}$.
Therefore,the only solution is $x = 0$.

(A) Let $\theta = \operatorname{cosec}^{-1} \frac{5}{3} + \tan^{-1} \frac{2}{3}$.
Since $\operatorname{cosec}^{-1} \frac{5}{3} = \sin^{-1} \frac{3}{5}$,we have $\theta = \sin^{-1} \frac{3}{5} + \tan^{-1} \frac{2}{3}$.
Let $\alpha = \sin^{-1} \frac{3}{5}$,then $\sin \alpha = \frac{3}{5}$,so $\tan \alpha = \frac{3}{4}$.
Thus,$\theta = \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3}$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$:
$\theta = \tan^{-1} \left( \frac{3/4 + 2/3}{1 - (3/4)(2/3)} \right) = \tan^{-1} \left( \frac{9/12 + 8/12}{1 - 6/12} \right) = \tan^{-1} \left( \frac{17/12}{6/12} \right) = \tan^{-1} \frac{17}{6}$.
The expression is $\cot \left( \frac{2019 \pi}{2} - \theta \right)$.
Since $\frac{2019 \pi}{2} = 1009 \pi + \frac{\pi}{2}$,we have $\cot \left( 1009 \pi + \frac{\pi}{2} - \theta \right) = \cot \left( \frac{\pi}{2} - \theta \right) = \tan \theta$.
Since $\theta = \tan^{-1} \frac{17}{6}$,$\tan \theta = \frac{17}{6}$.

(C) We know the trigonometric identities: $\sec ^2(\theta) = 1 + \tan ^2(\theta)$ and $\operatorname{cosec}^2(\theta) = 1 + \cot ^2(\theta)$.
Let $\theta_1 = \tan ^{-1} 3$,then $\tan(\theta_1) = 3$.
Let $\theta_2 = \cot ^{-1} 3$,then $\cot(\theta_2) = 3$.
Substituting these into the expression:
$\sec ^2(\tan ^{-1} 3) = 1 + \tan ^2(\tan ^{-1} 3) = 1 + (3)^2 = 1 + 9 = 10$.
$\operatorname{cosec}^2(\cot ^{-1} 3) = 1 + \cot ^2(\cot ^{-1} 3) = 1 + (3)^2 = 1 + 9 = 10$.
Adding these results together:
$10 + 10 = 20$.
Therefore,the correct option is $C$.

(C) Given equation is $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2}$.
For the domain,we must have $x(x+1) \ge 0$ and $0 \le x^2+x+1 \le 1$.
The condition $x^2+x+1 \le 1$ implies $x^2+x \le 0$,which means $x(x+1) \le 0$.
Combining $x(x+1) \ge 0$ and $x(x+1) \le 0$,we get $x(x+1) = 0$,so $x=0$ or $x=-1$.
If $x=0$,the equation becomes $\tan ^{-1} (0) + \sin ^{-1} (1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is a solution.
If $x=-1$,the equation becomes $\tan ^{-1} (0) + \sin ^{-1} (1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is also a solution.
Thus,there are $2$ real solutions.

(C) Given that $\cos ^{-1}(x)+\cos ^{-1}(y)+\cos ^{-1}(z)=3 \pi$.
We know that the range of $\cos ^{-1}(\theta)$ is $[0, \pi]$.
Therefore,the maximum value of each term $\cos ^{-1}(x)$,$\cos ^{-1}(y)$,and $\cos ^{-1}(z)$ is $\pi$.
For their sum to be $3 \pi$,each term must individually be equal to $\pi$.
Thus,$\cos ^{-1}(x)=\pi$,$\cos ^{-1}(y)=\pi$,and $\cos ^{-1}(z)=\pi$.
This implies $x = \cos(\pi) = -1$,$y = \cos(\pi) = -1$,and $z = \cos(\pi) = -1$.
Now,substitute these values into the expression $x(y+z)+y(z+x)+z(x+y)$:
$= (-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1)$
$= (-1)(-2) + (-1)(-2) + (-1)(-2)$
$= 2 + 2 + 2 = 6$.

(D) The general term of the series is $T_r = \tan^{-1}\left(\frac{1}{x^2 + (2r-1)x + (r^2-r+1)}\right)$ for $r=1, 2, \dots, n$.
We can rewrite the argument as:
$\frac{1}{1 + (x+r)(x+r-1)} = \frac{(x+r) - (x+r-1)}{1 + (x+r)(x+r-1)}$.
Thus,$T_r = \tan^{-1}(x+r) - \tan^{-1}(x+r-1)$.
Summing these terms from $r=1$ to $n$:
$y = \sum_{r=1}^{n} [\tan^{-1}(x+r) - \tan^{-1}(x+r-1)]$.
This is a telescoping series:
$y = (\tan^{-1}(x+1) - \tan^{-1}(x)) + (\tan^{-1}(x+2) - \tan^{-1}(x+1)) + \dots + (\tan^{-1}(x+n) - \tan^{-1}(x+n-1))$.
$y = \tan^{-1}(x+n) - \tan^{-1}(x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{1+(x+n)^2} - \frac{1}{1+x^2}$.
Evaluating at $x=0$:
$y'(0) = \frac{1}{1+n^2} - \frac{1}{1+0^2} = \frac{1}{1+n^2} - 1 = \frac{1 - 1 - n^2}{1+n^2} = -\frac{n^2}{1+n^2}$.

(A) The given series is $S = \sum_{r=1}^{\infty} \cot ^{-1}(2r^2) = \sum_{r=1}^{\infty} \tan ^{-1}\left(\frac{1}{2r^2}\right)$.
We can rewrite the argument of $\tan^{-1}$ as:
$\frac{1}{2r^2} = \frac{2}{4r^2} = \frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$,we get:
$S = \sum_{r=1}^{\infty} [\tan^{-1}(2r+1) - \tan^{-1}(2r-1)]$.
Expanding the summation:
$S = (\tan^{-1} 3 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 3) + (\tan^{-1} 7 - \tan^{-1} 5) + \dots$
This is a telescoping series where all intermediate terms cancel out.
$S = \lim_{n \to \infty} [\tan^{-1}(2n+1) - \tan^{-1}(1)]$.
Since $\lim_{n \to \infty} \tan^{-1}(2n+1) = \frac{\pi}{2}$ and $\tan^{-1}(1) = \frac{\pi}{4}$,
$S = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(A) Given that $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$.
Expanding the numerator on the left side: $\frac{x^{2}+2x+1}{x(x^{2}+1)} = \frac{x^{2}+1}{x(x^{2}+1)} + \frac{2x}{x(x^{2}+1)} = \frac{1}{x} + \frac{2}{x^{2}+1}$.
Comparing this with $\frac{A}{x} + \frac{Bx+C}{x^{2}+1}$,we get $A=1$,$B=0$,and $C=2$.
Now,we evaluate the expression $\operatorname{cosec}^{-1}\left(\frac{1}{A}\right)+\cot ^{-1}\left(\frac{1}{B}\right)+\sec ^{-1} C$.
Note that $\cot^{-1}(\frac{1}{0})$ is $\cot^{-1}(\infty) = 0$.
Substituting the values: $\operatorname{cosec}^{-1}(1) + \cot^{-1}(\infty) + \sec^{-1}(2) = \frac{\pi}{2} + 0 + \frac{\pi}{3} = \frac{5\pi}{6}$.

(B) For the quadratic equation $4^{\sec^2 \alpha} x^2 + 2x + (\beta^2 - \beta + \frac{1}{2}) = 0$ to have real roots,the discriminant $D$ must be $\geq 0$.
$D = (2)^2 - 4(4^{\sec^2 \alpha})(\beta^2 - \beta + \frac{1}{2}) \geq 0$
$4 - 4(4^{\sec^2 \alpha})(\beta^2 - \beta + \frac{1}{2}) \geq 0$
$4^{\sec^2 \alpha}(\beta^2 - \beta + \frac{1}{2}) \leq 1$
Since $\sec^2 \alpha \geq 1$,we have $4^{\sec^2 \alpha} \geq 4^1 = 4$.
Also,$\beta^2 - \beta + \frac{1}{2} = (\beta - \frac{1}{2})^2 + \frac{1}{4} \geq \frac{1}{4}$.
Multiplying these,$4^{\sec^2 \alpha}(\beta^2 - \beta + \frac{1}{2}) \geq 4 \times \frac{1}{4} = 1$.
For the inequality $4^{\sec^2 \alpha}(\beta^2 - \beta + \frac{1}{2}) \leq 1$ to hold,both expressions must be at their minimum values:
$4^{\sec^2 \alpha} = 4$ $\Rightarrow \sec^2 \alpha = 1$ $\Rightarrow \cos^2 \alpha = 1$.
$(\beta - \frac{1}{2})^2 = 0 \Rightarrow \beta = \frac{1}{2}$.
Therefore,$\cos^2 \alpha + \cos^{-1} \beta = 1 + \cos^{-1}(\frac{1}{2}) = 1 + \frac{\pi}{3}$.

(B) Let $\theta = 2 \tan^{-1} x$. Then $\cot \theta = \cot(2 \tan^{-1} x) = \frac{1 - x^2}{2x}$.
Given equation is $\sin(2 \cos^{-1}(\cot \theta)) = 0$.
This implies $2 \cos^{-1}(\cot \theta) = n\pi$ for some integer $n$.
So,$\cos^{-1}(\cot \theta) = \frac{n\pi}{2}$,which means $\cot \theta = \cos(\frac{n\pi}{2})$.
For the range of $\cos^{-1}$,we have $0 \le \cos^{-1}(\cot \theta) \le \pi$,so $n$ can be $0, 1, 2$.
Case $1$: $n=0 \implies \cos^{-1}(\cot \theta) = 0 \implies \cot \theta = 1 \implies \frac{1 - x^2}{2x} = 1 \implies x^2 + 2x - 1 = 0 \implies x = -1 \pm \sqrt{2}$.
Case $2$: $n=1 \implies \cos^{-1}(\cot \theta) = \frac{\pi}{2} \implies \cot \theta = 0 \implies \frac{1 - x^2}{2x} = 0 \implies x^2 = 1 \implies x = \pm 1$.
Case $3$: $n=2 \implies \cos^{-1}(\cot \theta) = \pi \implies \cot \theta = -1 \implies \frac{1 - x^2}{2x} = -1 \implies x^2 - 2x - 1 = 0 \implies x = 1 \pm \sqrt{2}$.
All these $6$ values are valid as they are in the domain of $\tan^{-1} x$ and satisfy the range of $\cos^{-1}$.
Thus,there are $6$ solutions.

(A) Let $x = \cos \theta$. As $x \rightarrow 1$,$\theta \rightarrow 0$.
Substituting this into the limit:
$\lim _{\theta \rightarrow 0} \frac{\sqrt{\cos \theta}-1}{\theta^2}$
Multiply the numerator and denominator by $(\sqrt{\cos \theta}+1)$:
$= \lim _{\theta \rightarrow 0} \frac{\cos \theta - 1}{\theta^2(\sqrt{\cos \theta}+1)}$
Using the identity $\cos \theta - 1 = -2 \sin^2(\frac{\theta}{2})$:
$= \lim _{\theta \rightarrow 0} \frac{-2 \sin^2(\frac{\theta}{2})}{\theta^2(\sqrt{\cos \theta}+1)}$
$= \lim _{\theta}$ ${\rightarrow 0} -2 \cdot \left(\frac{\sin(\frac{\theta}{2})}{\frac{\theta}{2} \cdot 2}\right)^2 \cdot \frac{1}{\sqrt{\cos \theta}+1}$
$= -2 \cdot \frac{1}{4} \cdot \frac{1}{1+1} = -2 \cdot \frac{1}{4} \cdot \frac{1}{2} = -\frac{1}{4}$.

(D) Given the mean of $a, b, 8, 5, 10$ is $6$:
$\frac{a+b+8+5+10}{5} = 6 \implies a+b+23 = 30 \implies a+b = 7$.
The variance is $6.80$:
$\frac{a^2+b^2+8^2+5^2+10^2}{5} - (6)^2 = 6.80$.
$\frac{a^2+b^2+64+25+100}{5} - 36 = 6.80$.
$\frac{a^2+b^2+189}{5} = 42.80$.
$a^2+b^2+189 = 214 \implies a^2+b^2 = 25$.
Since $(a+b)^2 = a^2+b^2+2ab$,we have $7^2 = 25 + 2ab \implies 49 = 25 + 2ab \implies 2ab = 24 \implies ab = 12$.
We need to find $\operatorname{Tan}^{-1} \frac{1}{a} + \operatorname{Tan}^{-1} \frac{1}{b} = \operatorname{Tan}^{-1} \left( \frac{\frac{1}{a} + \frac{1}{b}}{1 - \frac{1}{ab}} \right) = \operatorname{Tan}^{-1} \left( \frac{\frac{a+b}{ab}}{1 - \frac{1}{ab}} \right) = \operatorname{Tan}^{-1} \left( \frac{a+b}{ab-1} \right)$.
Substituting the values: $\operatorname{Tan}^{-1} \left( \frac{7}{12-1} \right) = \operatorname{Tan}^{-1} \frac{7}{11}$.

(A) Given equation: $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1}(x)$.
Taking $\sin$ on both sides:
$\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}} + \frac{4x}{5}\sqrt{1-\frac{9x^2}{25}} = x$.
If $x=0$,the equation holds true.
For $x \neq 0$,divide by $x$:
$\frac{3}{25}\sqrt{25-16x^2} + \frac{4}{25}\sqrt{25-9x^2} = 1$.
$3\sqrt{25-16x^2} + 4\sqrt{25-9x^2} = 25$.
Let $3\sqrt{25-16x^2} = 25 - 4\sqrt{25-9x^2}$.
Squaring both sides:
$9(25-16x^2) = 625 + 16(25-9x^2) - 200\sqrt{25-9x^2}$.
$225 - 144x^2 = 625 + 400 - 144x^2 - 200\sqrt{25-9x^2}$.
$200\sqrt{25-9x^2} = 800$.
$\sqrt{25-9x^2} = 4$.
$25-9x^2 = 16 \Rightarrow 9x^2 = 9 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.
Checking $x=1$: $\sin^{-1}(3/5) + \sin^{-1}(4/5) = \sin^{-1}(1) = \pi/2$. This is true.
Checking $x=-1$: $\sin^{-1}(-3/5) + \sin^{-1}(-4/5) = -(\sin^{-1}(3/5) + \sin^{-1}(4/5)) = -\pi/2 = \sin^{-1}(-1)$. This is true.
The values of $x$ are $0, 1, -1$. The sum is $0 + 1 + (-1) = 0$.

(B) Let $f(x) = 2(\cos ^{-1} x)^2-\pi \cos ^{-1} x+\frac{\pi^2}{4}$.
Let $y = \cos ^{-1} x$. Since $x \in [-1, 1]$,we have $y \in [0, \pi]$.
Then $f(y) = 2y^2 - \pi y + \frac{\pi^2}{4}$.
Completing the square: $f(y) = 2(y^2 - \frac{\pi}{2}y) + \frac{\pi^2}{4} = 2(y - \frac{\pi}{4})^2 - \frac{\pi^2}{8} + \frac{\pi^2}{4} = 2(y - \frac{\pi}{4})^2 + \frac{\pi^2}{8}$.
For $y \in [0, \pi]$,the minimum value occurs at $y = \frac{\pi}{4}$,which is $f(\frac{\pi}{4}) = \frac{\pi^2}{8}$.
The maximum value occurs at the boundary $y = \pi$ (since $|\pi - \frac{\pi}{4}| > |0 - \frac{\pi}{4}|$).
$f(\pi) = 2(\pi - \frac{\pi}{4})^2 + \frac{\pi^2}{8} = 2(\frac{3\pi}{4})^2 + \frac{\pi^2}{8} = 2(\frac{9\pi^2}{16}) + \frac{\pi^2}{8} = \frac{9\pi^2}{8} + \frac{\pi^2}{8} = \frac{10\pi^2}{8} = \frac{5\pi^2}{4}$.
The sum of the maximum and minimum values is $\frac{5\pi^2}{4} + \frac{\pi^2}{8} = \frac{10\pi^2 + \pi^2}{8} = \frac{11\pi^2}{8}$.

(D) We know that for $x \in [-1, 1]$,$\operatorname{Cos}^{-1}(-x) + \operatorname{Sin}^{-1}(-x) = \frac{\pi}{2}$.
Given the function $f(x) = \operatorname{Cos}^{-1}(-x) + \operatorname{Sin}^{-1}(-x) + \operatorname{Cosec}^{-1}(x)$,we substitute the identity:
$f(x) = \frac{\pi}{2} + \operatorname{Cosec}^{-1}(x)$.
The domain of $\operatorname{Cosec}^{-1}(x)$ is $(-\infty, -1] \cup [1, \infty)$.
Case $1$: If $x \in [1, \infty)$,then $x = 1$ is the only value in the domain of $\operatorname{Cos}^{-1}(-x)$ and $\operatorname{Sin}^{-1}(-x)$ (which is $[-1, 1]$). Thus,$x = 1$.
$f(1) = \frac{\pi}{2} + \operatorname{Cosec}^{-1}(1) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.
Case $2$: If $x \in (-\infty, -1]$,then $x = -1$ is the only value in the domain of $\operatorname{Cos}^{-1}(-x)$ and $\operatorname{Sin}^{-1}(-x)$. Thus,$x = -1$.
$f(-1) = \frac{\pi}{2} + \operatorname{Cosec}^{-1}(-1) = \frac{\pi}{2} - \frac{\pi}{2} = 0$.
Therefore,the range of the function is $\{0, \pi\}$.

(D) Let $u = x^2 + 2x + k$. The equation becomes $2 \operatorname{Cot}^{-1}(u) = \pi - 3 \operatorname{Tan}^{-1}(u)$.
Using the identity $\operatorname{Cot}^{-1}(u) = \frac{\pi}{2} - \operatorname{Tan}^{-1}(u)$,we substitute:
$2(\frac{\pi}{2} - \operatorname{Tan}^{-1}(u)) = \pi - 3 \operatorname{Tan}^{-1}(u)$
$\pi - 2 \operatorname{Tan}^{-1}(u) = \pi - 3 \operatorname{Tan}^{-1}(u)$
$\operatorname{Tan}^{-1}(u) = 0$,which implies $u = 0$.
So,$x^2 + 2x + k = 0$.
For this quadratic equation to have two distinct real solutions,the discriminant $D$ must be greater than $0$.
$D = b^2 - 4ac = (2)^2 - 4(1)(k) = 4 - 4k$.
Setting $D > 0$,we get $4 - 4k > 0$,which simplifies to $4 > 4k$,or $k < 1$.
Thus,$k \in (-\infty, 1)$.

(A) Let $\theta = \tan^{-1} \left(\frac{a}{\sqrt{b^2-a^2}}\right)$. Then $\tan \theta = \frac{a}{\sqrt{b^2-a^2}}$.
Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$,we have $\sec^2 \theta = 1 + \frac{a^2}{b^2-a^2} = \frac{b^2-a^2+a^2}{b^2-a^2} = \frac{b^2}{b^2-a^2}$.
Thus,$\sec \theta = \frac{b}{\sqrt{b^2-a^2}}$.
Now,let $\phi = \cos^{-1} x$. Then $\cos \phi = x$,which implies $\sin \phi = \sqrt{1-x^2}$.
So,$\cot \phi = \frac{\cos \phi}{\sin \phi} = \frac{x}{\sqrt{1-x^2}}$.
The given equation is $\cot \phi = \sec \theta$,so $\frac{x}{\sqrt{1-x^2}} = \frac{b}{\sqrt{b^2-a^2}}$.
Squaring both sides: $\frac{x^2}{1-x^2} = \frac{b^2}{b^2-a^2}$.
Cross-multiplying: $x^2(b^2-a^2) = b^2(1-x^2) = b^2 - b^2x^2$.
$x^2(b^2-a^2+b^2) = b^2$.
$x^2(2b^2-a^2) = b^2$.
$x^2 = \frac{b^2}{2b^2-a^2}$.
Taking the square root,$x = \frac{b}{\sqrt{2b^2-a^2}}$.

(A) Given,$x=\sin \left(2 \tan ^{-1} 2\right)$ and $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$.
For $x$,let $\tan ^{-1} 2 = \alpha$,so $\tan \alpha = 2$.
Then $x = \sin(2\alpha) = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} = \frac{2(2)}{1 + 2^2} = \frac{4}{5} = 0.8$.
For $y$,let $\tan ^{-1} \frac{4}{3} = \beta$,so $\tan \beta = \frac{4}{3}$.
Using $\cos \beta = \frac{1}{\sqrt{1 + \tan^2 \beta}} = \frac{1}{\sqrt{1 + (16/9)}} = \frac{1}{\sqrt{25/9}} = \frac{3}{5}$.
Then $y = \sin(\beta/2) = \sqrt{\frac{1 - \cos \beta}{2}} = \sqrt{\frac{1 - 3/5}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \approx 0.447$.
Since $0.8 > 0.447$,we have $x > y$.

(C) Given equation is $\sin [2 \cos^{-1} \cot (2 \tan^{-1} x)] = 0$.
Let $\theta = 2 \tan^{-1} x$. Then $\cot \theta = \cot (2 \tan^{-1} x) = \frac{1 - \tan^2(\tan^{-1} x)}{2 \tan(\tan^{-1} x)} = \frac{1 - x^2}{2x}$.
The equation becomes $\sin [2 \cos^{-1} (\frac{1 - x^2}{2x})] = 0$.
This implies $2 \cos^{-1} (\frac{1 - x^2}{2x}) = n\pi$ for some integer $n$.
So,$\cos^{-1} (\frac{1 - x^2}{2x}) = \frac{n\pi}{2}$.
Taking cosine on both sides,$\frac{1 - x^2}{2x} = \cos(\frac{n\pi}{2})$.
For the expression to be defined,the argument of $\cos^{-1}$ must be in $[-1, 1]$,so $|\frac{1 - x^2}{2x}| \le 1$.
Possible values for $\cos(\frac{n\pi}{2})$ are $0, 1, -1$.
Case $1$: $\frac{1 - x^2}{2x} = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.
Case $2$: $\frac{1 - x^2}{2x} = 1 \Rightarrow x^2 + 2x - 1 = 0 \Rightarrow x = -1 \pm \sqrt{2}$.
Case $3$: $\frac{1 - x^2}{2x} = -1 \Rightarrow x^2 - 2x - 1 = 0 \Rightarrow x = 1 \pm \sqrt{2}$.
All these $6$ values satisfy the condition $|\frac{1 - x^2}{2x}| \le 1$.
Thus,there are $6$ distinct values of $x$.

(B) Given equation: $\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$.
Let $\cos ^{-1} x = \theta$,then $x = \cos \theta$,where $\theta \in [0, \pi]$.
The equation becomes $\cos ^{-1}(1-\cos \theta) - 2\theta = \frac{\pi}{2}$.
$\cos ^{-1}(1-\cos \theta) = \frac{\pi}{2} + 2\theta$.
Taking $\cos$ on both sides:
$1-\cos \theta = \cos(\frac{\pi}{2} + 2\theta) = -\sin(2\theta)$.
$1-\cos \theta = -2\sin \theta \cos \theta$.
$1 - (1-2\sin^2(\frac{\theta}{2})) = -2(2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2}))\cos \theta$.
$2\sin^2(\frac{\theta}{2}) = -4\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})\cos \theta$.
Case $1$: $\sin(\frac{\theta}{2}) = 0 \implies \theta = 0 \implies x = \cos 0 = 1$.
Check $x=1$: $\cos ^{-1}(0) - 2\cos ^{-1}(1) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. This is a solution.
Case $2$: $\sin(\frac{\theta}{2}) = -2\cos(\frac{\theta}{2})\cos \theta$.
$\tan(\frac{\theta}{2}) = -2\cos \theta$. Since $\theta \in [0, \pi]$,$\tan(\frac{\theta}{2}) \ge 0$ and $-2\cos \theta$ can be negative. For $\theta \in [0, \pi/2]$,$\cos \theta \ge 0$,so $-2\cos \theta \le 0$. The only intersection is at $\theta=0$ (already found).
Thus,there is only one solution.

(A) Given expression: $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8} = k$.
Using $\cos(\frac{\pi}{2} + \theta) = -\sin \theta$,we have $\cos^4(\frac{\pi}{2} + \theta) = \sin^4 \theta$.
So,$\cos^4 \frac{5 \pi}{8} = \cos^4(\frac{\pi}{2} + \frac{\pi}{8}) = \sin^4 \frac{\pi}{8}$ and $\cos^4 \frac{7 \pi}{8} = \cos^4(\frac{\pi}{2} + \frac{3 \pi}{8}) = \sin^4 \frac{3 \pi}{8}$.
Substituting these,$k = (\cos^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8}) + (\cos^4 \frac{3 \pi}{8} + \sin^4 \frac{3 \pi}{8})$.
Using $a^2 + b^2 = (a+b)^2 - 2ab$,we get $k = [(\cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8})^2 - 2 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}] + [(\cos^2 \frac{3 \pi}{8} + \sin^2 \frac{3 \pi}{8})^2 - 2 \sin^2 \frac{3 \pi}{8} \cos^2 \frac{3 \pi}{8}]$.
Since $\sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2(2 \theta)$,$k = [1 - \frac{1}{2} \sin^2 \frac{\pi}{4}] + [1 - \frac{1}{2} \sin^2 \frac{3 \pi}{4}]$.
$k = 2 - \frac{1}{2}(\frac{1}{2}) - \frac{1}{2}(\frac{1}{2}) = 2 - \frac{1}{4} - \frac{1}{4} = 2 - \frac{1}{2} = \frac{3}{2}$.
Now,$\sin^{-1}(\sqrt{\frac{k}{2}}) + \cos^{-1}(\frac{k}{3}) = \sin^{-1}(\sqrt{\frac{3/2}{2}}) + \cos^{-1}(\frac{3/2}{3}) = \sin^{-1}(\frac{\sqrt{3}}{2}) + \cos^{-1}(\frac{1}{2})$.
$= \frac{\pi}{3} + \frac{\pi}{3} = \frac{2 \pi}{3}$.

(B) Given,$\theta = 2 \tan^{-1} \frac{1}{8} + 2 \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7}$.
Using the formula $2 \tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2}$,we have:
$2 \tan^{-1} \frac{1}{8} = \tan^{-1} \frac{2/8}{1-1/64} = \tan^{-1} \frac{1/4}{63/64} = \tan^{-1} \frac{16}{63}$.
$2 \tan^{-1} \frac{1}{5} = \tan^{-1} \frac{2/5}{1-1/25} = \tan^{-1} \frac{2/5}{24/25} = \tan^{-1} \frac{5}{12}$.
Now,$\theta = \tan^{-1} \frac{16}{63} + \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{1}{7}$.
Using $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x+y}{1-xy}$:
$\tan^{-1} \frac{16}{63} + \tan^{-1} \frac{5}{12} = \tan^{-1} \frac{16/63 + 5/12}{1 - (16/63)(5/12)} = \tan^{-1} \frac{(192+315)/756}{(756-80)/756} = \tan^{-1} \frac{507}{676} = \tan^{-1} \frac{3}{4}$.
So,$\theta = \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{1}{7} = \tan^{-1} \frac{3/4 + 1/7}{1 - (3/4)(1/7)} = \tan^{-1} \frac{25/28}{25/28} = \tan^{-1} (1) = \frac{\pi}{4}$.
Thus,$\tan \frac{\theta}{2} = \tan \frac{\pi}{8} = \sqrt{2} - 1$.
Given $\tan \frac{\theta}{2} = \sqrt{m} + \sqrt{n}$ with $m < n$,we have $\sqrt{m} + \sqrt{n} = -1 + \sqrt{2}$.
This implies $m = 1$ and $n = 2$ (since $m < n$).
Then $(m^n + n^m)^{m+n} = (1^2 + 2^1)^{1+2} = (1 + 2)^3 = 3^3 = 27$.

(D) We use the identities: $2 \tan^{-1} (\theta) = \sin^{-1} (\frac{2 \theta}{1 + \theta^2}) = \cos^{-1} (\frac{1 - \theta^2}{1 + \theta^2}) = \sec^{-1} (\frac{1 + \theta^2}{1 - \theta^2})$.
For $x = \sin (2 \tan^{-1} 2)$:
$x = \sin (\sin^{-1} (\frac{2 \times 2}{1 + 2^2})) = \frac{4}{5} = 0.8$.
For $y = \cos (2 \tan^{-1} 3)$:
$y = \cos (\cos^{-1} (\frac{1 - 3^2}{1 + 3^2})) = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8$.
For $z = \sec (2 \tan^{-1} 4)$:
$z = \sec (\sec^{-1} (\frac{1 + 4^2}{1 - 4^2})) = \frac{1 + 16}{1 - 16} = \frac{17}{-15} \approx -1.133$.
Comparing the values: $-1.133 < -0.8 < 0.8$,which means $z < y < x$.

(A) Given the expression: $\sum_{k=1}^n \tan^{-1} \left( \frac{1}{k^2+k+1} \right) = \tan^{-1} \theta$.
We know the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$.
We can rewrite the term inside the summation as: $\frac{1}{1+k(k+1)} = \frac{(k+1)-k}{1+k(k+1)}$.
Thus,$\tan^{-1} \left( \frac{1}{k^2+k+1} \right) = \tan^{-1} (k+1) - \tan^{-1} k$.
Now,the summation becomes a telescoping series:
$\sum_{k=1}^n (\tan^{-1} (k+1) - \tan^{-1} k) = (\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots + (\tan^{-1} (n+1) - \tan^{-1} n)$.
After canceling the intermediate terms,we are left with:
$\tan^{-1} (n+1) - \tan^{-1} 1 = \tan^{-1} \theta$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$:
$\tan^{-1} \left( \frac{(n+1)-1}{1+(n+1)(1)} \right) = \tan^{-1} \theta$.
$\tan^{-1} \left( \frac{n}{n+2} \right) = \tan^{-1} \theta$.
Therefore,$\theta = \frac{n}{n+2}$.

(C) Given equation: $\operatorname{Sin}^{-1} x - \operatorname{Cos}^{-1} x = \operatorname{Sin}^{-1}(3x - 2)$.
We know that $\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x = \frac{\pi}{2}$,so $\operatorname{Cos}^{-1} x = \frac{\pi}{2} - \operatorname{Sin}^{-1} x$.
Substituting this into the equation: $\operatorname{Sin}^{-1} x - (\frac{\pi}{2} - \operatorname{Sin}^{-1} x) = \operatorname{Sin}^{-1}(3x - 2)$.
$2\operatorname{Sin}^{-1} x - \frac{\pi}{2} = \operatorname{Sin}^{-1}(3x - 2)$.
Taking $\sin$ on both sides: $\sin(2\operatorname{Sin}^{-1} x - \frac{\pi}{2}) = 3x - 2$.
$-\cos(2\operatorname{Sin}^{-1} x) = 3x - 2$.
Using $\cos(2\theta) = 1 - 2\sin^2 \theta$,where $\theta = \operatorname{Sin}^{-1} x$,we get $\cos(2\operatorname{Sin}^{-1} x) = 1 - 2x^2$.
So,$-(1 - 2x^2) = 3x - 2$.
$2x^2 - 1 = 3x - 2 \implies 2x^2 - 3x + 1 = 0$.
$(2x - 1)(x - 1) = 0$.
Thus,$x = 1$ or $x = \frac{1}{2}$.
Check $x = 1$: $\operatorname{Sin}^{-1}(1) - \operatorname{Cos}^{-1}(1) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. $RHS$: $\operatorname{Sin}^{-1}(3(1) - 2) = \operatorname{Sin}^{-1}(1) = \frac{\pi}{2}$. So $x = 1$ is a solution.
Check $x = \frac{1}{2}$: $\operatorname{Sin}^{-1}(\frac{1}{2}) - \operatorname{Cos}^{-1}(\frac{1}{2}) = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6}$. $RHS$: $\operatorname{Sin}^{-1}(3(\frac{1}{2}) - 2) = \operatorname{Sin}^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$. So $x = \frac{1}{2}$ is a solution.
Given $\alpha > \beta$,we have $\alpha = 1$ and $\beta = \frac{1}{2}$.
Then $3\alpha + 4\beta = 3(1) + 4(\frac{1}{2}) = 3 + 2 = 5$.

(C) We know that $\cot ^{-1}(x) = \tan ^{-1}\left(\frac{1}{x}\right)$.
Thus,the given expression is $\lim _{n \rightarrow \infty} \sum_{r=1}^n \tan ^{-1}\left(\frac{1}{r^2+\frac{3}{4}}\right)$.
We can rewrite the argument as $\frac{1}{r^2+1-\frac{1}{4}} = \frac{1}{1+(r^2-\frac{1}{4})} = \frac{1}{1+(r-\frac{1}{2})(r+\frac{1}{2})}$.
Using the identity $\tan ^{-1}(a) - \tan ^{-1}(b) = \tan ^{-1}\left(\frac{a-b}{1+ab}\right)$,we have:
$\tan ^{-1}\left(\frac{(r+\frac{1}{2})-(r-\frac{1}{2})}{1+(r+\frac{1}{2})(r-\frac{1}{2})}\right) = \tan ^{-1}\left(r+\frac{1}{2}\right) - \tan ^{-1}\left(r-\frac{1}{2}\right)$.
Now,the sum becomes a telescoping series:
$S_n = \sum_{r=1}^n \left[\tan ^{-1}\left(r+\frac{1}{2}\right) - \tan ^{-1}\left(r-\frac{1}{2}\right)\right]$
$S_n = \left(\tan ^{-1} \frac{3}{2} - \tan ^{-1} \frac{1}{2}\right) + \left(\tan ^{-1} \frac{5}{2} - \tan ^{-1} \frac{3}{2}\right) + \dots + \left(\tan ^{-1} \left(n+\frac{1}{2}\right) - \tan ^{-1} \left(n-\frac{1}{2}\right)\right)$.
All intermediate terms cancel out,leaving $S_n = \tan ^{-1}\left(n+\frac{1}{2}\right) - \tan ^{-1}\left(\frac{1}{2}\right)$.
Taking the limit as $n \rightarrow \infty$:
$\lim _{n \rightarrow \infty} S_n = \tan ^{-1}(\infty) - \tan ^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2} - \tan ^{-1}\left(\frac{1}{2}\right) = \cot ^{-1}\left(\frac{1}{2}\right) = \tan ^{-1}(2)$.

(A) Given $y = \operatorname{Sin}^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$.
We know that $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$.
Substituting these,we get $y = \operatorname{Sin}^{-1}\left(\frac{|\cos \frac{x}{2} + \sin \frac{x}{2}| + |\cos \frac{x}{2} - \sin \frac{x}{2}|}{|\cos \frac{x}{2} + \sin \frac{x}{2}| - |\cos \frac{x}{2} - \sin \frac{x}{2}|}\right)$.
Given $\frac{-3\pi}{2} < x < \frac{-\pi}{2}$,we have $\frac{-3\pi}{4} < \frac{x}{2} < \frac{-\pi}{4}$.
In this interval,$\cos \frac{x}{2} < 0$ and $\sin \frac{x}{2} < 0$,and $|\sin \frac{x}{2}| > |\cos \frac{x}{2}|$.
Simplifying the expression inside the inverse sine,we get $y = \operatorname{Sin}^{-1}(\cot(\frac{x}{2}))$.
Since $\cot(\frac{x}{2}) = \tan(\frac{\pi}{2} - \frac{x}{2})$,we have $y = \operatorname{Sin}^{-1}(\tan(\frac{\pi}{2} - \frac{x}{2}))$. This simplifies to $y = \frac{\pi}{2} - (\frac{\pi}{2} - \frac{x}{2}) = \frac{x}{2}$ or similar depending on the branch.
However,differentiating $y = \frac{x}{2} + C$ gives $\frac{dy}{dx} = \frac{1}{2}$ or $-\frac{1}{2}$.
Given the constraints,the derivative is $-\frac{1}{2}$.

(D) Given equation: $\operatorname{Tan}^{-1} 1 + \frac{1}{2} \operatorname{Cos}^{-1} x^2 - \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right) = 0$.
Let $x^2 = \cos 2\theta$,where $2\theta \in [0, \pi]$. Then $\sqrt{1+x^2} = \sqrt{1+\cos 2\theta} = \sqrt{2}\cos\theta$ and $\sqrt{1-x^2} = \sqrt{1-\cos 2\theta} = \sqrt{2}\sin\theta$.
The third term becomes $\operatorname{Tan}^{-1}\left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right) = \operatorname{Tan}^{-1}\left(\tan(\frac{\pi}{4}+\theta)\right) = \frac{\pi}{4} + \theta$.
Substituting back: $\frac{\pi}{4} + \frac{1}{2}(2\theta) - (\frac{\pi}{4} + \theta) = 0$.
This simplifies to $0 = 0$,which is an identity.
However,the domain of the expression requires $1-x^2 \ge 0$,so $x^2 \le 1$,and the denominator $\sqrt{1+x^2}-\sqrt{1-x^2} \neq 0$,so $x^2 \neq 1$.
Also,$x^2 \ge 0$. Thus,$x^2 \in [0, 1)$.
Since $x^2 = \cos 2\theta$,this corresponds to $2\theta \in (0, \pi/2]$,which means $x^2 \in (0, 1]$.
Combining these,$x^2 \in (0, 1)$.
Since there are infinitely many values of $x$ in the interval $(-1, 0) \cup (0, 1)$,the number of solutions is infinitely many.

(D) Let $x = \cos \theta$. Then $f(x) = \operatorname{Sec}^{-1}\left(\frac{1}{2\cos^2 \theta - 1}\right) = \operatorname{Sec}^{-1}\left(\frac{1}{\cos 2\theta}\right) = \operatorname{Sec}^{-1}(\sec 2\theta) = 2\theta = 2\cos^{-1}x$.
Thus,$\frac{df}{dx} = 2 \times \left(-\frac{1}{\sqrt{1-x^2}}\right) = -\frac{2}{\sqrt{1-x^2}}$.
Now,let $x = \tan \phi$. Then $g(x) = \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+\tan^2 \phi}-1}{\tan \phi}\right) = \operatorname{Tan}^{-1}\left(\frac{\sec \phi - 1}{\tan \phi}\right) = \operatorname{Tan}^{-1}\left(\frac{1-\cos \phi}{\sin \phi}\right) = \operatorname{Tan}^{-1}\left(\tan \frac{\phi}{2}\right) = \frac{\phi}{2} = \frac{1}{2}\tan^{-1}x$.
Thus,$\frac{dg}{dx} = \frac{1}{2} \times \frac{1}{1+x^2} = \frac{1}{2(1+x^2)}$.
The derivative of $f(x)$ with respect to $g(x)$ is $\frac{df/dx}{dg/dx} = \frac{-2/\sqrt{1-x^2}}{1/(2(1+x^2))} = -\frac{4(1+x^2)}{\sqrt{1-x^2}}$.

(A) Given $f(x) = \cos(\tan^{-1}(\sin(\tan^{-1} x)))$.
Using $\tan^{-1} x = \sin^{-1} \frac{x}{\sqrt{1+x^2}}$,we have $\sin(\tan^{-1} x) = \frac{x}{\sqrt{1+x^2}}$.
Then ${f(x) = \cos(\tan^{-1}(\frac{x}{\sqrt{1+x^2}}}))$.
Using $\tan^{-1} \theta = \cos^{-1} \frac{1}{\sqrt{1+\theta^2}}$,we get $f(x) = \cos(\cos^{-1} \frac{1}{\sqrt{1+\frac{x^2}{1+x^2}}}) = \frac{1}{\sqrt{\frac{1+x^2+x^2}{1+x^2}}} = \sqrt{\frac{1+x^2}{1+2x^2}}$.
Now,$(f \circ f)(x) = f(f(x)) = \sqrt{\frac{1+(f(x))^2}{1+2(f(x))^2}}$.
Substituting $f(x)^2 = \frac{1+x^2}{1+2x^2}$,we get $(f \circ f)(x) = \sqrt{\frac{1+\frac{1+x^2}{1+2x^2}}{1+2(\frac{1+x^2}{1+2x^2})}} = \sqrt{\frac{1+2x^2+1+x^2}{1+2x^2+2+2x^2}} = \sqrt{\frac{2+3x^2}{3+4x^2}}$.
Taking the limit as $x \rightarrow \infty$,$\lim_{x \rightarrow \infty} \sqrt{\frac{2+3x^2}{3+4x^2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} = \frac{3}{2\sqrt{3}}$.

(A) The equation is given as,$y = \sin^{2} (\cot^{-1} \sqrt{\frac{1 + x}{1 - x}})$.
Substitute $x = \cos 2\theta$,then $\frac{1 + x}{1 - x} = \frac{1 + \cos 2\theta}{1 - \cos 2\theta} = \frac{2\cos^{2}\theta}{2\sin^{2}\theta} = \cot^{2}\theta$.
So,$y = \sin^{2} (\cot^{-1} \sqrt{\cot^{2}\theta}) = \sin^{2} (\cot^{-1} (\cot\theta)) = \sin^{2}\theta$.
Using the identity $\sin^{2}\theta = \frac{1 - \cos 2\theta}{2}$,we get $y = \frac{1 - x}{2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\frac{1}{2} - \frac{x}{2}) = 0 - \frac{1}{2} = -\frac{1}{2}$.

(C) Given $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}\right)$.
Let $x=a \tan \theta$,then $\theta=\tan ^{-1}\left(\frac{x}{a}\right)$.
Substituting $x$ in the expression:
$y=\cos ^{-1}\left(\frac{a^2-a^2 \tan ^2 \theta}{a^2+a^2 \tan ^2 \theta}\right) + \sin ^{-1}\left(\frac{2 a^2 \tan \theta}{a^2+a^2 \tan ^2 \theta}\right)$
$y=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) + \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$
Using trigonometric identities $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$ and $\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta}$:
$y=\cos ^{-1}(\cos 2 \theta) + \sin ^{-1}(\sin 2 \theta)$
$y=2 \theta + 2 \theta = 4 \theta$.
Substituting back $\theta = \tan ^{-1}\left(\frac{x}{a}\right)$:
$y=4 \tan ^{-1}\left(\frac{x}{a}\right)$.
Now,differentiating with respect to $x$:
$\frac{d y}{d x} = 4 \cdot \frac{1}{1+(\frac{x}{a})^2} \cdot \frac{1}{a}$
$\frac{d y}{d x} = 4 \cdot \frac{a^2}{a^2+x^2} \cdot \frac{1}{a} = \frac{4 a}{a^2+x^2}$.

(B) Given the quadratic equation $3x^2 - 16x + 5 = 0$.
Comparing with $ax^2 + bx + c = 0$,we have $a = 3, b = -16, c = 5$.
Sum of roots $\alpha + \beta = -\frac{b}{a} = -\left(\frac{-16}{3}\right) = \frac{16}{3}$.
Product of roots $\alpha \beta = \frac{c}{a} = \frac{5}{3}$.
Since $\alpha \beta = \frac{5}{3} > 1$,we use the formula:
$\tan^{-1} \alpha + \tan^{-1} \beta = \pi + \tan^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}\right)$.
Substituting this into the given expression:
$\left[\pi + \tan^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}\right)\right] - \tan^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}\right) = \pi$.

(C) Given $x+iy = \frac{1+7i}{(2-i)^2}$.
Expanding the denominator: $(2-i)^2 = 4 + i^2 - 4i = 4 - 1 - 4i = 3 - 4i$.
So,$x+iy = \frac{1+7i}{3-4i}$.
Multiplying the numerator and denominator by the conjugate $(3+4i)$:
$x+iy = \frac{(1+7i)(3+4i)}{(3-4i)(3+4i)} = \frac{3 + 4i + 21i + 28i^2}{9 - 16i^2} = \frac{3 + 25i - 28}{9 + 16} = \frac{-25 + 25i}{25} = -1 + i$.
Comparing real and imaginary parts,we get $x = -1$ and $y = 1$.
Now,evaluate $\operatorname{cosec}\left(\tan^{-1} \frac{y}{x} - \frac{\pi}{4}\right)$:
$= \operatorname{cosec}\left(\tan^{-1} \left(\frac{1}{-1}\right) - \frac{\pi}{4}\right) = \operatorname{cosec}\left(\tan^{-1}(-1) - \frac{\pi}{4}\right)$.
Since $\tan^{-1}(-1) = -\frac{\pi}{4}$,we have:
$= \operatorname{cosec}\left(-\frac{\pi}{4} - \frac{\pi}{4}\right) = \operatorname{cosec}\left(-\frac{\pi}{2}\right) = -1$.

(C) Given,$\sinh (2 \tanh ^{-1} x) = \frac{11}{60}$.
Using the identity $\sinh (2 \theta) = \frac{2 \tanh \theta}{1 - \tanh ^2 \theta}$,where $\theta = \tanh ^{-1} x$,we have $\tanh \theta = x$.
Substituting this into the equation:
$\frac{2x}{1 - x^2} = \frac{11}{60}$
$120x = 11(1 - x^2)$
$11x^2 + 120x - 11 = 0$
$11x^2 + 121x - x - 11 = 0$
$11x(x + 11) - 1(x + 11) = 0$
$(11x - 1)(x + 11) = 0$
Thus,$x = \frac{1}{11}$ or $x = -11$.
Since the domain of $\tanh ^{-1} x$ is $(-1, 1)$,$x = -11$ is rejected.
Therefore,$x = \frac{1}{11}$.

(A) We know that $\cos ^{-1} \left(\frac{3}{5}\right) = \tan ^{-1} \left(\frac{4}{3}\right)$ and $\sin ^{-1} \left(\frac{5}{13}\right) = \tan ^{-1} \left(\frac{5}{12}\right)$.
Substituting these into the expression:
$\tan ^{-1} \left(\frac{4}{3}\right) + \tan ^{-1} \left(\frac{5}{12}\right) + \tan ^{-1} \left(\frac{16}{63}\right)$
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left(\frac{x+y}{1-xy}\right)$ for the first two terms:
$= \tan ^{-1} \left[\frac{\frac{4}{3} + \frac{5}{12}}{1 - \left(\frac{4}{3} \times \frac{5}{12}\right)}\right] + \tan ^{-1} \left(\frac{16}{63}\right)$
$= \tan ^{-1} \left[\frac{\frac{16+5}{12}}{\frac{36-20}{36}}\right] + \tan ^{-1} \left(\frac{16}{63}\right) = \tan ^{-1} \left(\frac{21}{12} \times \frac{36}{16}\right) + \tan ^{-1} \left(\frac{16}{63}\right)$
$= \tan ^{-1} \left(\frac{63}{16}\right) + \tan ^{-1} \left(\frac{16}{63}\right)$
Since $\tan ^{-1} x + \tan ^{-1} \left(\frac{1}{x}\right) = \frac{\pi}{2}$ for $x > 0$,we get:
$= \frac{\pi}{2}$.

(C) We have $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$.
Since $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)=\cos ^{-1}\left(\frac{1}{3}\right)$,we have $\cos ^{-1}\left(\frac{1}{3}\right)+\sin ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}$. Thus,$A-V$.
$(B)$ Let $\theta=\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$. Then $\sin \theta=\frac{(-1)^n}{2}=\sin\left((-1)^n \frac{\pi}{6}\right)$. The general solution is $\theta=k \pi+(-1)^k\left((-1)^n \frac{\pi}{6}\right)$,which simplifies to $k \pi \pm \frac{\pi}{6}, k \in Z$. This matches $I$.
$(C)$ Let $\alpha=\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right) = \tan ^{-1}(\sqrt{2}+1)$. Since $\tan \frac{3\pi}{8} = \sqrt{2}+1$,we have $\alpha = \frac{3\pi}{8}$. Thus,$C-IV$.
$(D)$ Let $t=\sin ^{-1}|\sin x|$. Then $t=\sqrt{t} \Rightarrow t^2-t=0 \Rightarrow t(t-1)=0$. So $t=0$ or $t=1$. $\sin ^{-1}|\sin x|=0 \Rightarrow |\sin x|=0 \Rightarrow x=k\pi$. $\sin ^{-1}|\sin x|=1 \Rightarrow |\sin x|=\sin 1 \Rightarrow x=k\pi \pm 1$. Combining these,the general solution is $x=k\pi \pm 1, k \in Z$. Thus,$D-II$.